Suppose I have following list :
mat <- vector("list")
for (i in 1:4) mat[[i]] <- matrix(rnorm(25),5)
mat
> mat
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] -1.27171814 -0.8087277 -0.4771356 0.6001265 0.9842248
[2,] -1.37337445 1.0754536 -1.6304287 -0.6854075 -0.6029
Hi all,
Can anyone please guide me how to draw a Concentration ellipsoid for a
bivariate system with a bivariate normal dist. having a VCV matrix :
Sigma <- matrix(c(1,2,2,5), 2, 2)
I would like to draw in using GGPLOT. Your help will be highly appreciated.
Thanks,
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View this message in co
tion.
Any help please?
Thanks,
RON70 wrote:
>
> Hi all,
>
> Can anyone please guide me how to draw a Concentration ellipsoid for a
> bivariate system with a bivariate normal dist. having a VCV matrix :
>
> Sigma <- matrix(c(1,2,2,5), 2, 2)
>
> I would like to
Hi all,
Is there any process to conduct a search for a particular digit or letter in
a ch. string? For example I want to make a search where 1st numeric figure
appears in the string "asd123". Here the answer should be "4"
Thanks,
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Can I ask a small stat. related question here?
Suppose I have two predictors for a time series processes and accuracy of
predictor is measured from MSEs. My question is, if two predictors give same
MSE then, necessarily they have to be identical? Can anyone provide me any
counter example?
Thanks
Hi all,
Most of the time in my programming I use "for" loop however there will
always be some way to use other loop like "while", "if" etc to implement
same query. My question is among those different loops which is
theoretically fastest and why?
Thanks
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Hi, suppose I have following codes :
library(zoo); library(ggplot2)
dat <- matrix(rnorm(500*2), 500); dat <- zooreg(dat, start =
as.Date("01/01/01", "%m/%d/%y"), frequency=1); plot(dat)
head(dat); month.no <- format(index(dat), "%m"); dat1 <-
cbind(coredata(dat), as.numeric(month.no))
x <- dat1[
the plots I am getting is here :
http://www.2shared.com/file/6739681/d4c6c9d3/plot.html
and
http://www.2shared.com/file/6739673/bd50b430/plot.html
How you are getting transposed one?
smu-2 wrote:
>
> hey,
>
> On Sat, Jul 18, 2009 at 01:52:34AM -0700, RON70 wrote:
>>
>
;)$Version # devel version
> [1] "1.6-0"
>> R.version.string # Windows Vista
> [1] "R version 2.9.1 Patched (2009-07-16 r48939)"
>
> For the devel version of zoo see the installation instructions
> at the bottom of::
> http://r-forge.r-project.org/scm/?gr
0.8.2 reshape_0.8.2 plyr_0.1.9 proto_0.3-8
> zoo_1.5-5
> [6] mblm_0.11 chron_2.3-30lattice_0.17-25 Design_2.1-2
> survival_2.35-4
> [11] Hmisc_3.5-2
>
> loaded via a namespace (and not attached):
> [1] cluster_1.12.0 tools_2.9.1
> On Jul 18, 2009,
Hi,
I have 100 price data series like price1, price2, price3, . All
are "zoo" objects. Now I want to merge all them together. Obviously I can do
this using "merge(price1, price2, price3, )". However as I have lot
of price series (almost 1000) above systax is very tiresome. Is
I have following code on "qplot" :
library(ggplot2)
ggplot() + geom_histogram(aes(x=rnorm(100), fill=..count..), xlab="",
ylab="")
However above code doesnot seem to remove the xlab & ylab. What is the
correct code? I also want to remove the color-palate in the right side. Is
there any way to do
I am getting following strange error :
library(sn)
library(ggplot2)
dat1 <- as.matrix(rsn(1000, 0, 1, 0))[,1]
ggplot() + geom_histogram(aes(x = dat1, y = ..density.., fill = ..count..))
+
xlab("Distribution") + scale_y_continuous("")
Error: No data for layer
Can anyone please
In page http://had.co.nz/ggplot2/scale_date.html , there is a time series
plot for multiple variables at the bottom. i.e
qplot(date, value, data = em, geom = "line", group = variable) +
+ facet_grid(variable ~ ., scale = "free_y")
How can I make different color for different series?
Thanks
Thank you for your reply. Yes I am reading that document and ofcourse
simultaniously trying to apply in my current problem as well. Yes still I am
learning.
Here I have my code :
library(ggplot2)
library(plyr)
dat = rnorm(1000); variable = rep(c("Variable:1", "Variable:2"), each=500);
coll = rep
PLEASE IGNORE MY PREVIOUS MAIL
RON70 wrote:
>
> Thank you for your reply. Yes I am reading that document and ofcourse
> simultaniously trying to apply in my current problem as well. Yes still I
> am learning.
>
> Here I have my code :
>
> library(ggplot2)
> libra
Hi all, I was trying to draw a stacked density plot like that :
library(ggplot2); library(plyr)
dat <- cbind(rnorm(300), rep(c(1,2), each=150))
ggplot() + geom_density(aes(x=dat[,1], fill=factor(dat[,2]),
position="stack")) +
xlab("") + ylab("") +
scale_colour_manual(name = "
Please consider the following :
> expand.grid(rep(list(c(1:3)), 3))
Var1 Var2 Var3
1 111
2 211
3 311
4 121
5 221
6 321
7 131
8 231
9 331
10112
11212
1231
I have a vector of letters like c("a", "B", "c"). Is there any R function to
force all elements to lower-cap ?
Thanks,
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Sent from the R help mailing list archive at N
Hi all, is there any function to find some words in a character-string? For
example suppose the string is : "gdfsa-sdhchc-88", now I want to find
whether this string contains "sdhch". Is there any R function to do that?
Regards,
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s,
>
> --
> *Luc Villandré*
> /Biostatistician
> McGill University Health Center -
> Montreal Children's Hospital Research Institute/
>
> RON70 wrote:
>> Hi all, is there any function to find some words in a character-string?
>> For
>> example suppose the string is : &
Hi,
Suppose I have following dataset :
mat <- matrix(rnorm(100), 50)
Now I want to put 2nd column in the place of 1st and 1st column in the place
of 2nd. Is there any "quick" way to do that?
Thanks and regards,
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Hi all, Eigen vectors obtained from the function eigen() are ortho-normal? I
see the documentation however there is no formal mention on that. If no,
then is there any direct function to do the same?
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Dear all, is it possible to extract the seed that was used for some random
number generation? For example suppose I draw a random sample of size 1000
from a normal population :
rnorm(1000)
Now I want to know what seed R used for that sample generation. Any way out?
Thanks and regards,
--
View
Can anyone please explain me why it is not coming? I have done following :
> j=1
> i=0:j; sum( choose(i+j,i) )
[1] 3
> j=2
> i=0:j; sum( choose(i+j,i) )
[1] 10
> j=3
> i=0:j; sum( choose(i+j,i) )
[1] 35
> j=0:3; i=0:j; sum( choose(i+j,i) )
Warning message:
In 0:j : numerical expression has 4 elem
I have a dataframe with two columns :
11600 238'4
12000 218'0
12200 209'0
12600 192'0
13000 176'4
14000 145'0
15000 119'0
16000 103'0
1800080'0
1900068'3
2 59'0
11600 208'1
12000 189'2
12200 180'3
There are repeatations in 1st column and I want to use this as
I was trying to dw data from Economagic
[http://www.economagic.com/em-cgi/data.exe/libor/day-ussnon], using
following code :
library(fimport)
dat2 = economagicSeries("libor/day-ussnon", frequency = "daily")
Here I see that data is not complete, downloaded data starts from
"2007-12-31 ", whereas
Oh my mistake, I missed the argument nDaysBack = 366.
However pls let me know how to change the data to zoo object.
RON70 wrote:
>
> I was trying to dw data from Economagic
> [http://www.economagic.com/em-cgi/data.exe/libor/day-ussnon], using
> following code :
>
> libra
00
0.40.437500.443750.80625
2008-11-25 2008-11-26 2008-12-08 2008-12-10 2008-12-11 2008-12-12 2008-12-15
2008-12-16
0.931250.987500.186250.125000.115000.118750.11938
0.15938
Can anyone please tell me whether I missing something?
RON70 wrote:
>
ersion of zoo:
>
> source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/pkg/R/as.zoo.R?rev=557&root=zoo";)
> z <- as.zoo(dat2)
> time(z) <- as.Date(time(z)) # optional
>
>
>
> On Tue, Dec 30, 2008 at 6:25 AM, RON70 wrote:
>>
>
Hi, I want find all roots for the following polynomial :
a <- c(-0.07, 0.17); b <- c(1, -4); cc <- matrix(c(0.24, 0.00, -0.08,
-0.31), 2); d <- matrix(c(0, 0, -0.13, -0.37), 2); e <- matrix(c(0.2, 0,
-0.06, -0.34), 2)
A1 <- diag(2) + a %*% t(b) + cc; A2 <- -cc + d; A3 <- -d + e; A4 <- -e
fn <- fu
)
> lines(x1, pfunc(x1), col=2, lty=2)
>
> solve(p) # gives you the roots (some are, of course, complex)
>
>
> Hope this helps,
> Ravi.
>
>
>
> ____
>
> Ravi Varadhan, Ph.D.
> Assistant P
A "communication matrix" "K" is such that : K * VEC(A) = VEC(transpose of A).
Is there any readily available R function to find that Communication matrix?
Thanks
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Sent from the
I found there is serious difference between the result of ADF test in R and
Eviews, for this data :
> dat
V1
1 -0.075851693
2 -0.046125504
3 -0.009117161
4 0.025569817
5 0.034882743
6 0.073671497
7 0.063805297
8 0.062306796
9 0.072343820
10 0.058354121
11 -0.007635359
I am wondering how R calculate p-value for a test. Does R do some
"Approximate" integration on the p.d.f of null distribution? How I can see
the code for this particular calculation?
Your help will be highly appreciated.
Regards,
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Still no reply. Is my question not understandable at all?
RON70 wrote:
>
> I am wondering how R calculate p-value for a test. Does R do some
> "Approximate" integration on the p.d.f of null distribution? How I can see
> the code for this particular calculation?
>
&
Hi everyone,
I have this kind of raw dataset :
-
2005-01-17T00:00:00+05:30
10149
1288.40002
-
2005-01-18T00:00:00+05:30
10149
1291.69995
-
2005-01-19T00:00:00+05:30
10149
1288.19995
I was looking for some R procedure to extract data from this, that should
Sorry to be off-topic. Can somebody please explain me what is Portmanteau
test? Why it's name is like that? When I would say, a particular test is
portmanteau test? I did some googling but got no satisfactory answer at all.
Please anybody help for understanding that?
Regards,
--
View this messag
Still waiting for some input. Did my question void forum rule in any manner?
RON70 wrote:
>
> Sorry to be off-topic. Can somebody please explain me what is Portmanteau
> test? Why it's name is like that? When I would say, a particular test is
> portmanteau test? I did some go
Is there any function in R to clear values from all variables stored in
current and previous sessions? R site search do not give the desired result
at all, all results are about to clear screen only not variables.
Thanks,
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I want to draw following plot, given here
http://www.2shared.com/file/4327128/830b82c4/pic.html
for the following data :
dat <- cbind(rnorm(100), sample(c(1:4), 1000, T))
colnames(dat) <- c("data","level")
Here x-axis should be on "data" and y-axis is for "level" and z-axis should
be to display
I need to solve a equation like this :
a = b/(1+x) + c/(1+x)^2 + d/(1+x)^3
where a,b,c,d are known constant. Is there any R-way to do that?
Thanks in advance
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Sent from the R help m
92
> [1] 0.8554268
> [1] 0.829437
> [1] 0.8454056
> [1] 0.835527
> [1] 0.8416126
> [1] 0.837854
> [1] 0.8401717
> [1] 0.838741
> [1] 0.8396236
> [1] 0.839079
> [1] 0.839415
> [1] 0.8392076
> [1] 0.8393356
> [1] 0.8392566
> [1] 0.8393053
> [1] 0.8392753
> [1
using decompose() function how can I get only seasonally adjusted series?
GR-13 wrote:
>
> Here's something that may help you get started:
> library(ts)
> ?decompose
> ?stl
>
> Thanks,
> -Girish
>
> On Dec 5, 1:55 pm, Matthias <[EMAIL PROTECTED]> wrote:
>> Hi,
>> I?m looking for a package whi
mat more closely. Both the
> separator and the year format specs fail to match your input.
>
> > as.Date("10-02-2008", format = "%m/%d/%y")
> [1] NA
> > as.Date("10-02-2008", format = "%m-%d-%Y")
> [1] "2008-10-02"
>
>
I have a date-like-vector like :
> date_file
"10-02-2008" "10-03-2008" "10-06-2008" "10-07-2008" "10-09-2008"
"10-10-2008" "10-13-2008" "10-14-2008" "10-15-2008"
"10-16-2008" "10-17-2008" "10-20-2008" "10-21-2008" "10-22-2008"
"10-23-2008" "10-24-2008" "10-28-2008" "10-29-2008"
"10-30-2008" "1
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